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Author Question: Find an equation of the plane tangent to the surface xy3z2 = 2 at the point (2, 1, -1). (Read 66 times)

pragya sharda

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Question 1

Find an equation of the tangent plane to the surface  x2 + 2y2 + 3z2 = 6   at the point  (1, 1, 1).
◦ x + y + z = 3
◦ x + 2y + 3z = 6
◦ 3x + 2y + z = 6
◦ x + 4y + 9z = 14
◦ x + 2y - 3z = 6

Question 2

Find an equation of the plane tangent to the surface  xy3z2 = 2 at the point  (2, 1, -1).
◦ x + 5y - 4z = 11
◦ x = (2 + t) i + (1 + 6t) j + (-1 - 4t) k, t ∈  R
◦ x + 6y + 4z = 4
◦ x + 6y - 4z = 0
◦ x + 6y - 4z = 12


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Marked as best answer by pragya sharda on May 27, 2021

leahm14

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Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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pragya sharda

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Reply 2 on: May 27, 2021
YES! Correct, THANKS for helping me on my review


kjohnson

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Reply 3 on: Yesterday
Thanks for the timely response, appreciate it

 

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